Answer:
Light provides brightness to see and also light carries energy
Natural sources of light include our sun and other stars, where the source of energy is nuclear energy (recall that the moon does not produce light but merely reflects sunlight), lightning, where the source is electrical, and fire, where the energy source is chemical.
Answer:
Explanation:
90 rpm = 90 / 60 rps
= 1.5 rps
= 1.5 x 2π rad /s
angular velocity of flywheel
ω = 3π rad /s
Let I be the moment of inertia of flywheel
kinetic energy = (1/2) I ω²
(1/2) I ω² = 10⁷ J
I = 2 x 10⁷ / ω²
=2 x 10⁷ / (3π)²
= 2.2538 x 10⁵ kg m²
Let radius of wheel be R
I = 1/2 M R² , M is mass of flywheel
= 1/2 πR² x t x d x R² , t is thickness , d is density of wheel .
1/2 πR⁴ x t x d = 2.2538 x 10⁵
R⁴ = 2 x 2.2538 x 10⁵ / πt d
= 4.5076 x 10⁵ / 3.14 x .1 x 7800
= 184
R= 3.683 m .
diameter = 7.366 m .
b ) centripetal accn required
= ω² R
= 9π² x 3.683
= 326.816 m /s²
(1) You must find the point of equilibrium between the two forces,
<span>G * <span><span><span>MT</span><span>ms / </span></span><span>(R−x)^2 </span></span>= G * <span><span><span>ML</span><span>ms / </span></span><span>x^2
MT / (R-x)^2 = ML / x^2
So,
x = R * sqrt(ML * MT) - ML / (MT - ML)
R = is the distance between Earth and Moon.
</span></span></span>The result should be,
x = 3.83 * 10^7m
from the center of the Moon, and
R - x = 3.46*10^8 m
from the center of the Earth.
(2) As the distance from the center of the Earth is the number we found before,
d = R - x = 3.46*10^8m
The acceleration at this point is
g = G * MT / d^2
g = 3.33*10^-3 m/s^2
Answer:
400 W/m^2 and 31℃
Explanation:
The output heat flux q"= 20 W/m^2 (geven)
The output heat flux from.the wall to the air by convection
q"conv = h(ts - t∞)
q"conv = 20(50-30) = 400 W/m^2
Therefor, this case is unsteady and the wall temperature changes with time till the energy balance exist.
ENERGY BALANCE
The input energy must be equal to the output energy for steady state condition. If not the state will be unstaidy or transient.
2. Its noticed that the output heat flux is not that the I put heat flux, therefore the wall tempers will be decreased till the output heat flux is reduced to the value of the given input heat flux
T steady = T∞ +q"/h
= 30 + 20/20 = 31℃
